# Easy Construction of Bicentric Quadrilateral: What Is This About? A Mathematical Droodle

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Explanation ### Easy Construction of Bicentric Quadrilateral

A quadrilateral is bicentric if it's both inscriptable and circumscriptable. (Inscriptable means admitting an incircle. Circumscriptable means cyclic, i.e., admitting a circumcircle.) Bicentric quadrilaterals might seem exotic, but the applet shows how such quadrilaterals can be constructed easily. (Another construction appears elsewhere.)

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Let ABCD be a cyclic quadrilateral with vertices on a given circle w. Assume ABCD is also orthodiagonal, i.e., AC ⊥ BD. Then quadrilateral PQRS formed by the tangents to w at the vertices of ABCD is bicentric. (It's inscriptible by construction. The assertion therefore is about its being circumscriptible - cyclic.) The converse is also true: if PQRS is cyclic, then the diagonal of ABCD are orthogonal. Furthermore, if PQRS is cyclic, then the point of intersection E of the diagonals of ABCD, and the centers I and O of the two circles are collinear.

The latter assertion can be rephrased. We know that, the diagonals of a cyclic quadrilateral (ABCD in this case) and those of the quadrilateral formed by the points of tangency of ABCD and its incircle, are concurrent. Therefore we can say that, for a bicentric quadrilateral, the intersecition of the diagonals M, the incenter I and the circumcenter O are collinear.

### Proof

First of all note the following angle identities

 (1) ∠QCB = ∠BCQ = ∠BAC = ∠BDC = α. (2) ∠ADS = ∠DAS = ∠ABD = ∠ACD = β. (3) ∠AEB = ∠CED = γ.

We have to show that the diagonals of ABCD are orthogonal, i.e., γ = 90°, iff

 ∠PQR + ∠PSR = 180°,

or, which is the same, iff

 (4) ∠BQC + ∠ASD = 180°

Now, in ΔBCQ,

 ∠BQC + 2α = 180°,

so that from (1)

 (*1) ∠BQC + 2∠BAC = 180°,

 ∠ASD + 2β = 180°,

so that from (2)

 (*2) ∠ASD + 2∠ABD = 180°,

Thus, in ΔABE,

 γ = 180° - α - β = 180° - (180° - ∠BQC)/2 - (180° - ∠ASD)/2 = (∠BQC + ∠ASD)/2 = (∠PQR + ∠PSR)/2,

which proves the first part of the assertion. The fact that I, O, and E are collinear has been proven in [Dubrovsky] and [Honsberger]. An absolutely delicious proof appears as a consequence of another construction of bicentric quadrilaterals.

### References

1. G. Bennett, Bi-centric Quadrilaterals and the Pedal n-gon, in The Lighter Side of Mathematics, R.K.Guy and R.E.Woodrow, eds, MAA, 1994, p. 97
2. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971, p. 45
3. H. Dorrie, 100 Great Problems Of Elementary Mathematics, Dover Publications, NY, 1965, pp. 188-193
4. V. N. Dubrovsky, Solution to problem M1154, Kvant, n 8, 1989, pp 34-35 (in Russian), pdf is available at https://kvant.mccme.ru/1989/08/p34.htm.
5. R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 1960, p. 95
6. R. Honsberger, In Pólya's Footsteps, MAA, 1999, pp. 100-101  ### Various Geometric Constructions

• How to Construct Tangents from a Point to a Circle
• How to Construct a Radical Axis
• Constructions Related To An Inaccessible Point
• Inscribing a regular pentagon in a circle - and proving it
• The Many Ways to Construct a Triangle and additional triangle facts
• Easy Construction of Bicentric Quadrilateral II
• Star Construction of Shapes of Constant Width
• Four Construction Problems
• Geometric Construction with the Compass Alone
• Construction of n-gon from the midpoints of its sides
• Short Construction of the Geometric Mean
• Construction of a Polygon from Rotations and their Centers
• Squares Inscribed In a Triangle I
• Construction of a Cyclic Quadrilateral
• Circle of Apollonius
• Six Circles with Concurrent Pairwise Radical Axes
• Trisect Segment: 2 Circles, 4 Lines
• Tangent to Circle in Three Steps
• Regular Pentagon Construction by K. Knop
• 